# Real part of tail of logarithm

Given a positive integer $$n$$, consider $$f_n = -\min_{|z|=1} \Re \sum_{i>n} \frac{z^i e^{-i/n}}{i}$$. What can be said about the growth of $$f_n$$? How large can it get?

Taking maximum instead of minimum, the answer is obvious as the maximum is attained at $$z=1$$ and this becomes a real analysis problem (which in particular implies $$f_n$$ is bounded, but perhaps $$f_n$$ tends to $$0$$ at some rate).

An approach suggested by a friend is to express $$\sum_{i>n} t^i/i$$ in terms of a Cauchy integral related to $$-\ln (1-\zeta t)$$ and then perhaps apply a saddle point estimate (uniformly in $$|t|=e^{-1/n}$$) -- but I wasn't able to carry this out.

\begin{aligned} s_n(z)&:= \sum_{j>n} \frac{z^j e^{-j/n}}j \\ &=\sum_{j>n} z^j e^{-j/n} \int_0^\infty du\,e^{-ju} \\ &=\int_0^\infty du\,\sum_{j>n} z^j e^{-j/n}e^{-ju} \\ &=\frac{z^{n+1}}{e^{1+1/n}}\,\int_0^\infty du\,\frac{e^{-(n+1)u}}{1-z e^{-u-1/n}} \\ &=\frac{z^{n+1}}{e^{1+1/n}}\,\int_0^\infty dv\,\frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}. \end{aligned} \tag{1} Next, with $$n\ge1$$, $$|z|=1$$, and $$v>0$$, $$\begin{equation*} |1-z e^{-(v+1)/n}|\ge1-e^{-(v+1)/n}\ge1-\frac1{1+(v+1)/n}=\frac{v+1}{n+v+1}, \end{equation*}$$ whence $$\begin{equation*} \Big|\frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}\Big| \le2e^{-v}. \tag{2} \end{equation*}$$ Also, for all real $$t$$ and all real $$v>0$$, $$\begin{equation*} \Big|\frac{e^{-v}}{1+v-it}\Big|\le e^{-v}. \tag{3} \end{equation*}$$
Suppose now that $$n\to\infty$$ and $$z=e^{it/n}$$ for some real number $$t$$ possibly varying with $$n$$ but staying bounded. Then $$\begin{equation*} \frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}=\frac{e^{-v}}{1+v-it}+o(1) \end{equation*}$$ for each real $$v>0$$. So, by (1), (2), (3) and dominated convergence, \begin{align*} s_n(z)&= \frac{e^{it}}e\,\int_0^\infty dv\,\frac{e^{-v}}{1+v-it}+o(1)=\Gamma(0,1-i t)+o(1). \end{align*}
On the other hand, if for $$z$$ with $$|z|=1$$ it is not true that $$z=e^{it/n}$$ for some real number $$t$$ possibly varying with $$n$$ but staying bounded, then without loss of generality $$n(1-z)\to\infty$$, whence $$\begin{equation*} n|1-z e^{-(v+1)/n}|\ge n|1-z|-(v+1)\to\infty \end{equation*}$$ for each real $$v>0$$. So, by (1), (2), and dominated convergence, $$|s_n(z)|\to0$$.
So, $$\begin{equation*} -f_n=\min_{|z|=1} \Re s_n(z)\to\mu, \end{equation*}$$ where $$\begin{equation*} \mu:=\min_{t>0}r(t),\quad r(t):=\Re\Gamma(0,1-i t). \end{equation*}$$ (According to Mathematica's numerics, $$\mu=r(t_*)=-0.12686\dots$$, where $$t_*=2.0287\dots$$.)